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Question 1 Report
Question 3 Report
Answer Details
Question 4 Report
R = 14.14N
R = F = ma
14.14 = 2 x a
a = 7.07ms-1
Question 5 Report
Question 6 Report
Two rays of light from a point below the surface of water are equally inclined to each other at \(60^\circ\) in water. What is the angle between the rays when they emerge into air? (Take the refractive index of water to be \(\frac{4}{3}\))
r1 = 41.8o and r2 = 41.8o
angle between them = r1 + r2
= 83.6o
Question 7 Report
= m(v2−u2)2s
F = 0.05(0−(500)2)2×0.25
F = 25 000N
Question 8 Report
x = (0 x 10) + (12 x 10 x 10 x 10) = 500m
distance in 9s; x - (0 x 9) + (12 x 10 x 9 x 9) = 405m
distance in last is = difference
= 500 - 405
= 95m
Question 9 Report
Answer Details
Question 10 Report
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Question 12 Report
V1 = 2V2
1.5×2V2−73+273 = 4.5×V2T2
T2 = 300K = 27oC
Question 13 Report
= Height = 1031.5×102
= 6.67mm
Question 14 Report
2 X 10-5 x 2.5 = A2×1.51.5×(−20−20)
A2 = 1.498m2
Question 16 Report
Answer Details
Question 17 Report
R.d = height insideH2Oheight inside liquid
0.8 = 23X
Question 18 Report
depth = velocity x time
= 1500 x 0.5
= 750m = 0.75km
Question 19 Report
Question 20 Report
Answer Details
Question 21 Report
In the Fig above, Current I passes through the combination, if the power dissipated in the 5 ohm resistor is 40W, then the power dissipated in the 10 ohm resistor is
Answer Details
Question 22 Report
1.98gs = 1.98 x 1
= 600 cou.
Q = It
600 = 2 x t
t = 300sec = 5 mins
Question 23 Report
Answer Details
Question 25 Report
F x d2 = constant
F1 x (d1)2 = F2 x (d2)2
d2 = 2d1
F1 x (d1)2 = F2 x (2d1)2
F2 = 14 F1
Question 26 Report
load5 = 1000.5
load = 1000N
Question 27 Report
The graph in the fig above describes the motion of a particle. the acceleration of the particle during the motion is
Answer Details
Question 28 Report
RQRP = LQAPLPAQ
LQ = 2LP
DQ = 2DP
AP = π (2DP)22
= π DP
AQ = π (DP)22
= 14 π DP
therefore, RQRP = 2LP×πDPLP×14πDP
= 2 : 1
Question 29 Report
Two thermo flasks of volume \(V_x\) and \(V_y\) are filled with liquid water at an initial temperature of 0oC. After sometime the temperature were found to be \(\theta_x\), \(\theta_y\), respectively. Given that \(\frac{V_x}{V_y}=2\) and \(\frac{\theta_x}{\theta_y}=\frac{1}{2}\) the ratio of the heat flow into the flask is
Hx = Dx x Vx x Cx x θ x
Hy = Dy x Vy x Cy x θ y
HxHy = Dx×Vx×Cx×θxDx×Vx×Cx×θx
= 2 x 12
= 1
Question 30 Report
Answer Details
Question 31 Report
5002500 = 120Vs
Vs = 600v
Question 32 Report
Question 33 Report
Question 34 Report
Two divers G and H are at depths 20m and 40m respectively below the water surface in a lake. The pressure on G is P1 while the pressure on H is P2. If the atmospheric pressure is equivalent to 10m of water, then the value of \( \frac{p_2}{p_1} \) is
p1p1 = 4020
= 2
Question 35 Report
2T2 = √L12L1
T2 = 2.83sec
Question 36 Report
A cell gives a current of 0.15A through a resistance of \(8\ \Omega\) and 0.3A. When the resistance is changed to \(3\ \Omega\) the internal resistance of the cell is
Answer Details
Question 37 Report
Answer Details
Question 38 Report
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Question 39 Report
In the Fig above, MN is a light uniform meter rule pivoted at O, the 80cm mark. A load of mass 3.00kg is suspended on the meter rule at L, the 10cm mark. If the rule is kept in equilibrium by a string RP, fixed at P and attached to the rule at R, the 20cm mark, then the Tension T on the string is
Answer Details
Question 40 Report
= Nm3 x m31
= Nm = work
Question 41 Report
e1 = 20 - L ; e2 = 5cm
f1e1 = f2e2
10020−L = 1005
hence, L = 15cm
Question 42 Report
What is the number of neutrons in the Uranium isotope \( {}^{238}_{92}X \)?
=146
Question 43 Report
= 340170
= 2m ; for closed pipe
L = 14 λ
= 14 x 21
= 0.5m
= 50cm
Question 44 Report
6 x 105 = V2R t
R = 240×240×5×606×105
= 28.8Ω
Question 46 Report
Question 47 Report
Question 49 Report
m x (80 - θ ) x 1.5 = m x (θ - 20) x 1
θ = 56oC
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